Lintcode: Unique Paths

C++

dp

递推式：dp[i][j] = dp[i-1][j] + dp[i][j-1]

初值：dp[i][j] = 1，i=0 or j=0

空间优化：省掉一维

 class Solution {
 public:
     /**
      * @param n, m: positive integer (1 <= n ,m <= 100)
      * @return an integer
      */
     int uniquePaths(int m, int n) {
         // wirte your code here
         vector<vector<int> > dp(m,vector<int>(n));
         for (int i = ; i < m ; ++i) {
             for (int j = ; j < n; ++j) {
                 if ( i ==  ) {
                     dp[i][j] = ;
                 } else if ( j ==  ) {
                     dp[i][j] = ;
                 } else {
                     dp[i][j] = dp[i - ][j] + dp[i][j - ];
                 }
             }
         }
         return dp[m - ][n - ];
     }
 };

空间优化

 class Solution {
 public:
     /**
      * @param n, m: positive integer (1 <= n ,m <= 100)
      * @return an integer
      */
     int uniquePaths(int m, int n) {
         // wirte your code here
         // vector<vector<int> > dp(m,vector<int>(n));
         vector<int> dp(n);
         for (int i = ; i < m ; ++i) {
             for (int j = ; j < n; ++j) {
                 if ( i ==  ) {
                     dp[j] = ;
                 } else if ( j ==  ) {
                     dp[j] = ;
                 } else {
                     dp[j] = dp[j] + dp[j - ];
                 }
             }
         }
         return dp[n - ];
     }
 };